Does there exist a continuous function of $f : \mathbb R^2 \longrightarrow \mathbb R$ such that it is continuous whose both the partial derivatives don't exist.

I think the function $f : \mathbb R^2 \longrightarrow \mathbb R$ defined by $f(x,y) = |x|(1 + y)$, where $(x,y) \in \mathbb R^2$ has the above property at $(0,0)$. But I can't prove that $f$ is continuous at $(0,0)$ by $\epsilon-\delta$ method. Please help me.

Thank you in advance.

  • $\begingroup$ Do you must to prove with the $\;\epsilon-\delta\;$ method? What about $\;||x|(1+y)|\le2|x|\xrightarrow{}0\;$ ? $\endgroup$ – DonAntonio Dec 31 '16 at 17:21
  • $\begingroup$ Do you know that the product of continuous functions is continuous? And that $f(x,y)=|x|$ and $f(x,y)=1+y$ are continuous? It's usually easier to use those properties than $\varepsilon$-$\delta$. $\endgroup$ – Milo Brandt Dec 31 '16 at 17:21

Just consider the sum of two one-variable Weierstrass functions, one in the $x$ variable, one in the $y$ variable. This is even better, it has continuity everywhere and no differentiability or partial derivatives anywhere.


Let $\epsilon>0$ given.

we have to find $\delta>0$ such that:

$|x-0|<\delta $and $|y-0|<\delta \implies ||x|(1+y)|<\epsilon$.

as we are near $(0,0)$, we can assume that $\color{red}{-0.5}<y<\color{red}{0.5}$. which gives $0.5<1+y<1.5$ and $|1+y|<1.5$.

It is now easier to look for $\delta.\;:$ $|f(x,y)-f(0,0)|=|x||1+y|<1.5|x|<\epsilon$

or $|x|<\frac{2\epsilon}{3}$. thus, we will take $$\delta=\min(\frac{2\epsilon}{3},\color{red}{0.5})$$ to satisfy both conditions

  • $|y|<0.5$

  • $|x|<\frac{2\epsilon}{3}$.


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